Problem 11 - Calculus for Biology and Medicine

Question

Let \(\left(X_{1}, X_{2}, \ldots, X_{n}\right)\) denote a sample of size \(n\). Show that $$ \sum_{k=1}^{n}\left(X_{k}-\bar{X}\right)=0 $$ where \(\bar{X}\) is the sample mean.

Step-by-Step Solution

Answer

The sum of deviations from the sample mean is zero because they cancel each other out.

Step 1: Understand the Problem

We need to show that the sum of the deviations of each sample element from the sample mean is zero. Mathematically, this is expressed as \( \sum_{k=1}^{n} \left(X_{k} - \bar{X}\right) = 0 \).

Step 2: Recall the Definition of Sample Mean

The sample mean \( \bar{X} \) is defined as the sum of the sample elements divided by the number of elements \( n \): \( \bar{X} = \frac{1}{n} \sum_{k=1}^{n} X_{k} \).

Step 3: Express the Sum of Deviations

Substitute the sample mean \( \bar{X} \) into the expression for the sum of deviations: \( \sum_{k=1}^{n} \left(X_{k} - \bar{X}\right) = \sum_{k=1}^{n} X_{k} - \sum_{k=1}^{n} \bar{X} \).

Step 4: Simplify the Sum of Sample Mean Deviations

Since \( \bar{X} \) is a constant, it can be factored out of the sum: \( \sum_{k=1}^{n} \bar{X} = n \bar{X} \). Therefore, the expression becomes \( \sum_{k=1}^{n} X_{k} - n \bar{X} \).

Step 5: Substitute the Sample Mean

Replace \( n \bar{X} \) with \( \sum_{k=1}^{n} X_{k} \) because they are equal by the definition of the sample mean. This gives \( \sum_{k=1}^{n} X_{k} - \sum_{k=1}^{n} X_{k} = 0 \).

Step 6: Conclude the Proof

Having simplified the expression, we see that it reduces to zero, thus proving the initial statement.

Key Concepts

Deviations

When we talk about deviations in statistics, we refer to the differences between individual data points and a measure of central tendency, often the mean. In our case, we have a sample of size \( n \), and each data point in this sample is represented as \( X_k \), where \( k \) varies from 1 to \( n \). The sample mean, denoted as \( \bar{X} \), is considered the center of our sample data. To understand deviations, consider every data point \( X_k \) and see how much it deviates from \( \bar{X} \). Mathematically, this deviation for each point is expressed as \( X_k - \bar{X} \). When we sum up all these individual deviations over the entire sample, we get:

• \( \sum_{k=1}^{n} (X_k - \bar{X}) \)

This sum of deviations helps assess the spread or dispersion of data around the mean. An important revelation is that this sum of deviations is always zero, if calculated using the sample mean. This occurs because the total of positive and negative deviations balances each other, anchoring the data points around the mean.

Proof

Proving that the sum of deviations from the sample mean is zero is both straightforward and illuminating. Let's go through how this proof unfolds.First, we start with the definition of the sample mean:

• \( \bar{X} = \frac{1}{n} \sum_{k=1}^{n} X_k \)

This equation states that \( \bar{X} \) is the average of all sample values.Next, calculate the sum of deviations:

• \( \sum_{k=1}^{n} (X_k - \bar{X}) = \sum_{k=1}^{n} X_k - \sum_{k=1}^{n} \bar{X} \)

Rewriting the second term, because \( \bar{X} \) is just a constant repeated \( n \) times, we simplify using:

• \( \sum_{k=1}^{n} \bar{X} = n\bar{X} \)

By substituting \( n\bar{X} \) with \( \sum_{k=1}^{n} X_k \) (from our sample mean definition), the expression simplifies to:

• \( \sum_{k=1}^{n} X_k - n\bar{X} = \sum_{k=1}^{n} X_k - \sum_{k=1}^{n} X_k = 0 \)

The result, as expected, is zero, confirming our earlier statement.

Mathematical Expression

Mathematical expressions are pivotal in understanding statistical concepts. By manipulating symbols and numbers, we can derive meaningful insights. In the context of this exercise, we deal with the expression related to deviations and summarize its transformation.The core mathematical expression we aim to evaluate is the sum of deviations:

• \( \sum_{k=1}^{n} (X_k - \bar{X}) \)

Substituting \( \bar{X} \) using its definition, \( \bar{X} = \frac{1}{n} \sum_{k=1}^{n} X_k \), allows us to transform our expression:

• \( \sum_{k=1}^{n} X_k - \sum_{k=1}^{n} \bar{X} \)

This transformation is important as it reveals how the components interconnect to ultimately yield zero when simplified. Understanding that \( \sum_{k=1}^{n} \bar{X} \) can be rewritten as \( n\bar{X} \) is crucial. This insight is grounded firmly in the essence of the sample mean, which distributes equally across each sample observation:

• \( n\bar{X} = \sum_{k=1}^{n} X_k \)

Therefore, our mathematical journey shows the elegance and coherence in fundamental statistics, illustrating the concept as the sum of deviations equates to zero.